A proton of mass ' m_P ' has same energy as that of a photon of wavelength ' \lambda '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

Dual Nature of Radiation PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q58 with Solution

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Correct answer: D

For a photon of wavelength $\lambda$ ,

E = \frac{hc}{\lambda}

So,

\lambda = \frac{hc}{E}

The proton has the same energy $E$ and is moving non-relativistically, so its kinetic energy is

E = \frac{p^2}{2m_p}

Hence,

p = \sqrt{2m_pE}

Using de Broglie relation,

\lambda_p = \frac{h}{p} = \frac{h}{\sqrt{2m_pE}}

We need

\frac{\lambda_p}{\lambda}

Substitute $\lambda_p$ and $\lambda$ :

\frac{\lambda_p}{\lambda} = \frac{\dfrac{h}{\sqrt{2m_pE}}}{\dfrac{hc}{E}}

Cancel $h$ :

\frac{\lambda_p}{\lambda} = \frac{E}{c\sqrt{2m_pE}}

= \frac{1}{c}\sqrt{\frac{E}{2m_p}}

This matches Option D:

\boxed{\frac{1}{c}\sqrt{\frac{E}{2m_p}}}